# Is it possible to list $\mathbb{Q}$ so that the result set to be a monotonic sequence?

Let $\mathbb{Q}$ be the set of rational numbers. Is it possible, relabeling if needed, to list $\mathbb{Q}$ such that the result set to be a monotonic sequence? If not, why? If it is true, where is $\frac{r_1+r_2}{2}?$

This question occurs to me when I am reading Apostol's book, Mathematical Analysis, Page 68, Exercise 3.42. That exercise is : Consider the metric space ℚ of rational numbers with the Euclidean metric of ℝ. Let S consist of all rational numbers in the open interval (a,b), where a and b are irrational. Then S is is a closed and bounded subset of ℚ which is not compact.

When was ready to prove that $S$ is not compact, I wanted to construct an open covering of $S$ whose any finite subcovering can not cover $S$. Then I asked myself the above question. I feel that the answer is negative, but I can not find any explanation. Can anyone help me?

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## migrated from mathoverflow.netApr 10 '14 at 3:18

This question came from our site for professional mathematicians.

Good question, but not appropriate for mathoverflow, where we do research. – Gerry Myerson Apr 10 '14 at 0:40
@GerryMyerson, I am very sorry for my improper question. Actually, this question occurs to me when I was doing an exercise in Apostol's book, Mathematical Analysis, Page 68, Exercise 3.42. That exercise is : Consider the metric space $\mathbb{Q}$ of rational numbers with the Euclidean metric of $\mathbb{R}.$ Let $S$ consist of all rational numbers in the open interval $(a,b),$ where $a$ and $b$ are irrational. Then $S$ is is a closed and bounded subset of $\mathbb{Q}$ which is not compact. Maybe my question is not a research. – nuage Apr 10 '14 at 1:00

Assume such a listing exists. Consider two successive terms $a_n$ and $a_{n+1}$. Then the number $\frac12(a_n+ a_{n+1})$ whether it appears after the $(n+1)$th term or before the $n$th term it will violate monotonicity. So, it is not possible.
Regarding the finite subcover question, you don't necessarily need such an ordering of the rationals if you just choose the rational endpoints to your open cover appropriately. For example, take $U_{n}=(a_{n},b_{n})\cap\mathbb{Q}$ for all $n\in\mathbb{N}$, where $$a_{n}\in (a+\frac{1}{n+1},a+\frac{1}{n})\cap\mathbb{Q}$$ and $$b_{n}\in (b-\frac{1}{n},b-\frac{1}{n+1})\cap\mathbb{Q}.$$ If $a$ and $b$ are close to each other you can start labelling these sets from a large $n\in\mathbb{N}$ to ensure you get subsets of $S$. Note that $\mathbb{Q}$ is dense in $\mathbb{R}$ so each $a_{n}$ and $b_{n}$ exists, and by definition of the subspace topology each $U_{n}$ is open in $\mathbb{Q}$. Now $(a_{n})_{n=1}^{\infty}$ is a strictly decreasing sequence of rational numbers with $$\lim_{n\to\infty}a_{n}=a$$ and $(b_{n})_{n=1}^{\infty}$ is a strictly increasing sequence of rational numbers with $$\lim_{n\to\infty}b_{n}=b.$$ You can use the sandwich theorem for example to conclude that. Hence $S=\bigcup_{n=1}^{\infty}U_{n}$, and $(U_{n})_{n=1}^{\infty}$ is a strictly increasing sequence, i.e. $U_{n}\subset U_{n+1}$ for all $n\in\mathbb{N}$. What happens if you take a finite subcover? There will exist $N\in\mathbb{N}$ for instance so that this subcover fails to cover all $q\in (b-\frac{1}{N},b)\cap\mathbb{Q}\subseteq S$.
Assume it is possible. Restrict to an interval I . Then, e.g., by Bolzano-Weirstass' thm., there is a limit point 'a' of the sequence of Rationals in I, and all-but-finitely-many Rationals in the interval are in $(a-r,a+r)$, and then there are subintervals $I' \subset I$ with no Rationals in them.